If the pairs of lines `x^2+2x y+a y^2=0` and `a x^2+2x y+y^2=0` have exactly one line in common, then the joint equation of the other two lines is given by (a)`3x^2+8x y-3y^2=0` (b)`3x^2+10 x y+3y^2=0` (c) `y^2+2x y-3x^2=0` (d) `x^2+2x y-3y^2=0`

Options

A. `3x^(2)+8xy-3y^(2)=0`
B. `3x^(2)+10xy+3y^(2)=0`
C. `y^(2)+2xy-3x^(2)=0`
D. `x^(2)+2xy-3y^(2)=0`

Answer

B

Explanation

Let `y=mx` be a line common to the given pairs of lines . Then,
`am^(2)+2m+1=0andm^(2)+2m+a=0`
`:. (m^(2))/(2(1-a))=(m)/(a^(2)-1)=(1)/(2(1-a))`
or `m^(2)=1andm=-(a+1)/(2)`
or `(a+1)^(2)=4`
or `a=1or-3`
But for `a=1` , the two pairs have both the lines common . So `a=-3` and the slope m of the line common to both the pairs is 1. now ,
`:.x^(2)+2xy+ay^(2)=x^(2)+2xy-3y^(2)=(x-y)(x+3y)`
and `ax^(2)+2xy+y^(2)=-3x^(2)+2xy+y^(2)=-(x-y)(3x+y)`
So , the equation of the required pair of lines is
`(x+3y)(3x+y)=0`
or `3x^(2)+10xy+3y^(2)=0`